1/(x^2+x+1)^2的不定积分怎么求？ 求详细步骤

Question

如题所述

Answer 1

∫ 1/(x²+x+1)² dx
=∫ 1/[(x+1/2)²+3/4]² dx
令t=x+1/2,
dt=dx
=∫ 1/(t²+3/4)² dt
令t=√3/2*tan s,
dt=√3/2*sec²s ds
=√3/2*∫ sec²s/(3/4*tan²s+3/4)² ds
=8/(3√3)*∫ cos²s ds
=8/(3√3)*1/2*∫ (1+cos2s) ds
=4/(3√3)*∫ dp+4/(3√3)*∫ cos2s ds
=4/(3√3)*s+4/(3√3)*1/2*sin2s+C
=4/(3√3)*s+4/(3√3)*sinscoss+C
=4/(3√3)*arctan(2t/√3)+4/(3√3)*2t/√(4t²+3)*√3/√(4t²+3)+C
=4/(3√3)*arctan(2t/√3)+4/(3√3)*(2√3*t)/(4t²+3)+C
=4/(3√3)*arctan[(2/√3)*(x+1/2)]+4/(3√3)*2√3*(x+1/2)/[4(x+1/2)²+3]+C
=4/(3√3)*arctan[(2x+1)/√3]+4/(3√3)*(2√3*x+√3)/[4(x²+x+1)]+C
=4/(3√3)arctan[(2x+1)/√3]+(1/3)(2x+1)/(x²+x+1)+C